Evaluating $\int \frac{dx}{x^2 - 2x} dx$ $$\int \frac{dx}{x^2 - 2x}$$
I know that I have to complete the square so the problem becomes.
$$\int \frac{dx}{(x - 1)^2 -1}dx$$
Then I set up my A B and C stuff
$$\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{-1}$$
With that I find $A = -1, B = -1$ and $C = 0$ which I know is wrong.
I must be setting up the $A, B, C$ thing wrong but I do not know why.
 A: Added: "I know that I have to complete the square" is ambiguous. I interpreted it as meaning that the OP thought that completing the square was necessary to solve the problem. 
Completing the square is not a universal tool.  To find the integral efficiently, you certainly do not need to complete the square.
The simplest approach is to use partial fractions.  The bottom factors as $x(x-2)$. Find numbers $A$ and $B$ such that 
$$\frac{1}{x^2-2x}=\frac{A}{x-2}+\frac{B}{x}.$$
A: 
My book is telling me that I have to complete the square

$$I=\begin{eqnarray*}
\int \frac{dx}{x^{2}-2x} &=&\int \frac{dx}{\left( x-1\right) ^{2}-1}\overset{
u=x-1}{=}\int \frac{1}{u^{2}-1}\,du=-\text{arctanh }u+C
\end{eqnarray*},$$
$$\tag{1}$$
where I have used the substitution $u=x-1$ and the standard derivative $$\frac{d}{du}\text {arctanh}=\frac{1}{1-u^{2}}\tag{2}$$
You just need to substitute $u=x-1$ to write $\text{arctanh }u$ in terms of $x$.
Added 2: Remark. If we use the logarithmic representation of the inverse hyperbolic function $\text{arctanh }u$
$$\begin{equation*}
\text{arctanh }u=\frac{1}{2}\ln \left( u+1\right) -\frac{1}{2}\ln \left(
1-u\right),\qquad (\text{real for }|u|<1)\tag{3} 
\end{equation*}$$
we get for $u=x−1 $
$$\begin{eqnarray*}
I &=&-\text{arctanh }u+C=-\text{arctanh }\left( x-1\right) +C \\
&=&-\frac{1}{2}\ln x+\frac{1}{2}\ln \left( 2-x\right) +C \\
&=&\frac{1}{2}\left( \ln \frac{2-x}{x}\right) +C\qquad (0<x<2).
\end{eqnarray*}\tag{4}$$
Added. If your book does require using partial fractions then you can proceed as follows
$$\begin{equation*}
\int \frac{1}{u^{2}-1}\,du=\int \frac{1}{\left( u-1\right) \left( u+1\right) 
}\,du=\int \frac{1}{2\left( u-1\right) }-\frac{1}{2\left( u+1\right) }du.
\end{equation*}$$
$$\tag{5}$$
A: $$\begin{align} & {} \quad \int \frac{dx}{x^2 - 2x}\\ &=\int \frac{dx}{(x - 1)^2 -1}dx\\ &=\int \frac{dx}{(x - 1-1) (x-1+1)} \\ &=\int \frac{dx}{x(x-2)}\end{align}$$
The rest is easy (partial fractions):
$$\frac 12 \left[\int \frac {dx}{x-2} - \int \frac {dx}x\right]=\frac 12 \ln|x-2| -\frac 12 \ln |x| +C$$
A: I suppose that you could have also used trigonometric substitutions. Completing the square as you have done, we get that 
$$\displaystyle \int \frac{dx}{x^2-2x} = \int \frac{dx}{(x-1)^2 - 1}.$$
We now let $x-1 = \sec \theta.$ Notice then that $dx = \tan \theta \sec \theta  \ d\theta.$ We now have 
$$ \int \frac{dx}{(x-1)^2 - 1} = \int \frac{\tan \theta \sec \theta \ d\theta}{\sec^2\theta-1} = \int \frac{\tan \theta \sec \theta \ d\theta}{\tan^2 \theta} = \int \frac{\sec \theta \ d\theta}{\tan \theta} = \int \csc \theta \ d\theta. $$
And,
$$ \int \csc \theta \ d\theta = \ln (\csc \theta - \cot \theta) + C = \ln \left(\frac{\sqrt{x-2}}{\sqrt{x}} \right) + C,$$ as desired. 
(This is because from $x-1 = \sec \theta,$  we get $1/(x-1) = \cos \theta$ and thus $\sin \theta = (\sqrt{x^2-2x})/(x-1)$, and we just make the necessary substitutions in $\csc \theta - \cot \theta$).
